b
Frequency at \(1000 \,nm =\frac{3 \times 10}{1000 \times 10^{-9}} \Rightarrow 3 \times 10^{14} \,Hz\) available for channel band width \(=\frac{2}{100} \times 3 \times 10^{14} \Rightarrow 6 \times 10^{12} \,Hz\)
Bandwidth for \(1\) channel \(=8000 \,Hz\)
\(\therefore\) No. of channel
\(=\frac{6 \times 10^{12}}{8 \times 10^{3}} \Rightarrow \frac{600}{8} \times 10^{7}=75 \times 10^{7}\)